Let a n n = 0 ∞ be a sequence such that a 0 = a 1 = 0 and a n + 2 = 2 a n + 1 - a n + 1 for all n…

Given $a_{n+2}=2a_{n+1}-a_n+1$

Dividing by $7^{n+2}$, we get

$\frac{a_{n+2}}{7^{n+2}}=\frac{2}{7}·\frac{a_{n+1}}{7^{n+1}}-\frac{1}{49}·\frac{a_n}{7^n}+\frac{1}{7^{n+2}}$

So, $\sum _{n=2}^{\infty }\frac{a_{n+2}}{7^{n+2}}=\frac{2}{7}\sum _{n=2}^{\infty }\frac{a_{n+1}}{7^{n+1}}-\frac{1}{49}\sum _{n=2}^{\infty }\frac{a_n}{7^n}+\sum _{n=2}^{\infty }\frac{1}{7^{n+2}}$

Let
$\sum _{n=2}^{\infty }\frac{a_n}{7^n}=P$

$\Rightarrow \left(P-\frac{a_3}{7^3}-\frac{a_2}{7^2}\right)=\frac{2}{7}\left(P-\frac{a_2}{7^2}\right)-\frac{1}{49}P+\frac{1}{7^4}\left(\frac{{\displaystyle 1}}{{\displaystyle 1-\frac{1}{7}}}\right)$

$\left(\because a_2=1 \& a_3=3\right)$

$\Rightarrow P\left(1-\frac{2}{7}+\frac{1}{49}\right)=\frac{3}{7^3}+\frac{1}{7^2}-\frac{2}{7^3}+\frac{1}{6·7^3}$

$\Rightarrow \frac{36P}{49}=\frac{1}{42}$ $\Rightarrow P=\frac{7}{216}$